Question: $f(x, y, z) = \sin(z + x) - x^2 - y^2$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( 0, 0, \dfrac{\pi}{2} + \pi k \right)$, where $k = \ldots -1, 0, 1 \ldots$ (Choice B) B $\left( 0, 0, \pi k \right)$, where $k = \ldots -1, 0, 1 \ldots$ (Choice C) C $(0, 0, 0)$ (Choice D) D There are no critical points.
A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} \cos(z + x) -2x \\ \\ -2y \\ \\ \cos(z + x) \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} \cos(z + x) -2x = 0 \\ \\ -2y = 0 \\ \\ \cos(z + x) = 0 \end{cases}$ The equation $-2y = 0$ implies that $y$ must equal zero. If we subtract the equation $\cos(z + x) = 0$ from the equation $\cos(z + x) - 2x = 0$, we get the new equation $-2x = 0$. This tells us that $x$ must also equal zero. Now we can find what values of $z$ make the gradient equal to zero. Substituting $x = 0$, we get $\cos(z) = 0$. There are many values for $z$ that satisfy this equation. For example, $z = \dfrac{\pi}{2}$ makes $\cos(z) = 0$. Note that $\cos(z)$ is generally periodic every $2\pi$, but it crosses the $x$ -axis every $\pi$. So we can add any integer multiple of $\pi$ to get a new solution. All solutions will have the form $z = \dfrac{\pi}{2} + \pi k$, for some integer $k$. Therefore, $f$ has critical points at $\left( 0, 0, \dfrac{\pi}{2} + \pi k \right)$, where $k = \ldots -1, 0, 1 \ldots$.